JIPMER Jipmer Medical Solved Paper-2009

  • question_answer
    A \[0.1\] aqueous solution of a weak acid is 2% ionised. If the ionic product of water is \[1\times {{10}^{-4}},\] the \[[O{{H}^{-}}]\] is

    A)  \[5\times {{10}^{-12}}M\]                          

    B)  \[2\times {{10}^{-3}}M\]            

    C)  \[1\times {{10}^{-14}}M\]          

    D)         None of these

    Correct Answer: A

    Solution :

    Since, it is a weak acid, the equation to calculate \[[{{H}^{+}}]\]is \[=C\times \alpha \] (\[\alpha \] = % of ionisation) \[=0.1\times 0.02=2\times {{10}^{-3}}M\]         \[{{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]\] \[[O{{H}^{-}}]=\frac{{{K}_{w}}}{[{{H}^{+}}]}=\frac{1\times {{10}^{-14}}}{2\times {{10}^{-3}}}\] \[=5\times {{10}^{-12}}M\]


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